If $\lim_{x\to \infty}xf(x^2+1)=2$ then find $\lim_{x\to 0}\dfrac{2f'(1/x)}{x\sqrt{x}}$ 
If $\lim_{x\to \infty}xf(x^2+1)=2$ then find 
  $$\lim_{x\to 0}\dfrac{2f'(1/x)}{x\sqrt{x}}=?$$

My Try :
$$g(x):=xf(x^2+1)\\g'(x)=f(x^2+1)+2xf'(x^2+1)$$
Now what?
 A: Call $x^2+1=1/t$. Then
$$
\lim_{t\to 0^+}\sqrt{1/t-1}f(1/t)=\lim_{t\to 0^+}\frac{f(1/t)}{(1/t-1)^{-1/2}}=2.
$$
Using L'Hopital, if the following limit exists (ratio of derivatives upstairs and downstairs)
$$
\lim_{t\to 0^+}\frac{(-1/t^2)f'(1/t)}{\frac{1}{2 \left(\frac{1}{t}-1\right)^{3/2} t^2}}=\lim_{t\to 0^+} -2\left(\frac{1}{t}-1\right)^{3/2}f'(1/t)
$$
should also be equal to $2$. But this is equal to
$$
\lim_{t\to 0^+} -2\frac{f'(1/t)}{t^{3/2}}\ ,
$$
since $(1-t)^{3/2}\to 1$. Therefore the result you are after is $-2$. 
A: Note that, by letting $t=x^2+1$, we have
$$2=\lim_{x\to \infty}xf(x^2+1)=\lim_{t\to +\infty}\sqrt{t-1}f(t)=\lim_{t\to +\infty}\sqrt{t}f(t).$$
Hence, if the desired limit exists, then, by letting $t=1/x$,
\begin{align}
\lim_{x\to 0^+}\frac{2f'(1/x)}{x\sqrt{x}}&=
\lim_{t\to+\infty}2t\sqrt{t}f'(t)
=\lim_{t+\infty}t^{2}f(t)f'(t)\\
&=-\frac{1}{2}\lim_{t\to +\infty}\frac{(f^2(t))'}{(1/t)'}=-\frac{1}{2}\lim_{t\to +\infty}\frac{f^2(t)}{1/t}\\&=-\frac{1}{2}\left(\lim_{t\to+\infty}\sqrt{t}f(t)\right)^2=-\frac{4}{2}=-2
\end{align}
where we used L'Hôpital's rule in the second line.
